Pandas 在行上设置多索引，然后转置到列
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Pandas setting multi-index on rows, then transposing to columns
提问by sheridp
If I have a simple dataframe:
如果我有一个简单的数据框：
print(a)
one two three
0 A 1 a
1 A 2 b
2 B 1 c
3 B 2 d
4 C 1 e
5 C 2 f
I can easily create a multi-index on the rows by issuing:
我可以通过发出以下命令轻松地在行上创建多索引：
a.set_index(['one', 'two'])
three
one two
A 1 a
2 b
B 1 c
2 d
C 1 e
2 f
Is there a similarly easy way to create a multi-index on the columns?
是否有类似的简单方法在列上创建多索引？
I'd like to end up with:
我想结束：
one A B C
two 1 2 1 2 1 2
0 a b c d e f
In this case, it would be pretty simple to create the row multi-index and then transpose it, but in other examples, I'll be wanting to create a multi-index on both the rows and columns.
在这种情况下，创建行多索引然后转置它会非常简单，但在其他示例中，我将要在行和列上创建多索引。
采纳答案by piRSquared
Yes! It's called transposition.
是的！这叫做转位。
a.set_index(['one', 'two']).T
Let's borrow from @ragesz's post because they used a much better example to demonstrate with.
让我们借用@ragesz 的帖子，因为他们使用了一个更好的例子来演示。
df = pd.DataFrame({'a':['foo_0', 'bar_0', 1, 2, 3], 'b':['foo_0', 'bar_1', 11, 12, 13],
'c':['foo_1', 'bar_0', 21, 22, 23], 'd':['foo_1', 'bar_1', 31, 32, 33]})
df.T.set_index([0, 1]).T
回答by Nickil Maveli
You could use pivot_tablefollowed by a series of manipulations on the dataframe to get the desired form:
您可以使用pivot_table后跟对数据框的一系列操作来获得所需的形式：
df_pivot = pd.pivot_table(df, index=['one', 'two'], values='three', aggfunc=np.sum)
def rename_duplicates(old_list): # Replace duplicates in the index with an empty string
seen = {}
for x in old_list:
if x in seen:
seen[x] += 1
yield " "
else:
seen[x] = 0
yield x
col_group = df_pivot.unstack().stack().reset_index(level=-1)
col_group.index = rename_duplicates(col_group.index.tolist())
col_group.index.name = df_pivot.index.names[0]
col_group.T
one A B C
two 1 2 1 2 1 2
0 a b c d e f
回答by ragesz
I think the short answer is NO. To have multi-index columns, the dataframe should have two (or more) rows to be converted into headers (like columns for multi-index rows). If you have this kind of dataframe, creating multi-index header is not so difficult. It can be done in a very long line of code, and you can reuse it at any other dataframe, only the row numbers of the headers should be kept in mind & change if differs:
我认为简短的回答是否定的。要拥有多索引列，数据框应该有两（或更多）行要转换为标题（如多索引行的列）。如果您有这种数据帧，创建多索引标头就不是那么困难了。它可以在很长的代码行中完成，您可以在任何其他数据帧中重用它，只应记住标题的行号并在不同时更改：
df = pd.DataFrame({'a':['foo_0', 'bar_0', 1, 2, 3], 'b':['foo_0', 'bar_1', 11, 12, 13],
'c':['foo_1', 'bar_0', 21, 22, 23], 'd':['foo_1', 'bar_1', 31, 32, 33]})
The dataframe:
数据框：
a b c d
0 foo_0 foo_0 foo_1 foo_1
1 bar_0 bar_1 bar_0 bar_1
2 1 11 21 31
3 2 12 22 32
4 3 13 23 33
Creating multi-index object:
创建多索引对象：
arrays = [df.iloc[0].tolist(), df.iloc[1].tolist()]
tuples = list(zip(*arrays))
index = pd.MultiIndex.from_tuples(tuples, names=['first', 'second'])
df.columns = index
Multi-index header result:
多索引头结果：
first foo_0 foo_1
second bar_0 bar_1 bar_0 bar_1
0 foo_0 foo_0 foo_1 foo_1
1 bar_0 bar_1 bar_0 bar_1
2 1 11 21 31
3 2 12 22 32
4 3 13 23 33
Finally we need to drop 0-1 rows then reset the row index:
最后我们需要删除 0-1 行然后重置行索引：
df = df.iloc[2:].reset_index(drop=True)
The "one-line" version (only thing you have to change is to specify header indexes and the dataframe itself):
“一行”版本（您唯一需要更改的是指定标头索引和数据帧本身）：
idx_first_header = 0
idx_second_header = 1
df.columns = pd.MultiIndex.from_tuples(list(zip(*[df.iloc[idx_first_header].tolist(),
df.iloc[idx_second_header].tolist()])), names=['first', 'second'])
df = df.drop([idx_first_header, idx_second_header], axis=0).reset_index(drop=True)
